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Question Number 197171 by MM42 last updated on 09/Sep/23

$$2^{{log}_3(x^2+1)}+2\times {(x^2+1)}^{{log}_{3}2}=12$$$$\Rightarrow x=?$$$$$$

Answered by Frix last updated on 09/Sep/23

$$a^{{\log }_{b}c}=c^{{\log }_{b}a}$$$$\mathrm{Which}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{prove}:$$$$\ln \left(a^{{\log }_{b}c}\right)={\log }_{b}c\ln a=\frac{\ln a\ln c}{\ln b}=$$$$={\log }_{b}a\ln c=\ln \left(c^{{\log }_{b}a}\right)$$$$\mathrm{Here}\mathrm{we}\mathrm{have}$$$$2^{{\log }_3(x^2+1)}={(x^2+1)}^{\frac{\ln 2}{\ln 3}}$$$$\mathrm{The}\mathrm{equation}\mathrm{now}\mathrm{is}$$$$3{(x^2+1)}^{\frac{\ln 2}{\ln 3}}=12$$$${(x^2+1)}^{\frac{\ln 2}{\ln 3}}=4$$$$x^2+1=4^{\frac{\ln 3}{\ln 2}}=9$$$$x^2=8$$$$x=\pm 2\sqrt{2}$$

Commented by MM42 last updated on 10/Sep/23

$$verygood.\cancel{⋚}$$

Commented by Frix last updated on 10/Sep/23

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